if-M-is-a-point-on-the-line-y-x-and-points-P-0-1-Q-2-0-are-such-that-PM-PQ-is-minimum-then-find-P-

Question Number 148993 by gsk2684 last updated on 02/Aug/21

i f M i s a p o i n t o n t h e l i n e y = x a n d

p o i n t s P (0, 1), Q (2, 0) a r e s u c h t h a t

P M + P Q i s m i n i m u m t h e n f i n d P

Commented by mr W last updated on 02/Aug/21

i t h i n k t h e q u e s t i o n s h o u l d b e

f i n d M s u c h t h a t P M + Q M i s

m i n i m u m .

Commented by gsk2684 last updated on 02/Aug/21

y e s

t y p o g r a p h i c a l e r r o r

s o l u t i o n p l e a s e

Commented by bramlexs22 last updated on 02/Aug/21

if PM + MQ minimum \Rightarrow M ?

let f (x) = \sqrt{x^{2} + {(x - 1)}^{2}} + \sqrt{{(x - 2)}^{2} + x^{2}}

f {(x)}_{\min} if points M, P and Q colinear

\Rightarrow so we get m_{PQ} = m_{MQ}

\Rightarrow \frac{1 - 0}{0 - 2} = \frac{x - 0}{x - 2}

\Rightarrow - \frac{1}{2} = \frac{x}{x - 2}

\Rightarrow - x + 2 = 2 x; x = \frac{2}{3}

thus M (\frac{2}{3}, \frac{2}{3})

Commented by bramlexs22 last updated on 02/Aug/21

Commented by gsk2684 last updated on 04/Aug/21

i h a v e f o u n d t h a t f e e t o f p e r p e n d i c u l a r s

f r o m g i v e n p o i n t s o n t o t h e g i v e n

l i n e

a n d u s i n g t h e s e r i g h t a n g l e d t r i a n g l e s

c a n i g o f u r t h e r t o f i n d M s u c h t h a t

P M + Q M t o b e m i n i m u m

h e l p m e

Answered by iloveisrael last updated on 02/Aug/21

M (x, x) \Rightarrow PM + PQ = \sqrt{x^{2} + {(x - 1)}^{2}} + \sqrt{2^{2} + {(- 1)}^{2}}

= \sqrt{5} + \sqrt{{2 x}^{2} - 2 x + 1}

minimum when 4 x - 2 = 0; x = \frac{1}{2}

then M (\frac{1}{2}, \frac{1}{2})

why P ?

Commented by gsk2684 last updated on 02/Aug/21

t h a n k y o u

Answered by Kamel last updated on 02/Aug/21

i f M i s a p o i n t o n t h e l i n e y = x a n d

p o i n t s P (0, 1), Q (2, 0) a r e s u c h t h a t

P M + P Q i s m i n i m u m t h e n f i n d P

P M + Q M m i n i m u m \Rightarrow M \in (P Q) / M (x, x)

(P Q) : y = a x + 1 \Rightarrow (P Q) : y = - \frac{1}{2} x + 1

M \in (P Q) \Leftrightarrow x = y = - \frac{1}{2} x + 1 \Rightarrow x = y = \frac{2}{3}

Facebook Tweet Pin